Problem: You have found the following ages (in years) of all 5 turtles at your local zoo: $ 103,\enspace 2,\enspace 1,\enspace 79,\enspace 25$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{103 + 2 + 1 + 79 + 25}{{5}} = {42\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $103$ years $61$ years $3721$ years $^2$ $2$ years $-40$ years $1600$ years $^2$ $1$ year $-41$ years $1681$ years $^2$ $79$ years $37$ years $1369$ years $^2$ $25$ years $-17$ years $289$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{3721} + {1600} + {1681} + {1369} + {289}} {{5}} $ $ {\sigma^2} = \dfrac{{8660}}{{5}} = {1732\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{1732\text{ years}^2}} = {41.6\text{ years}} $ The average turtle at the zoo is 42 years old. There is a standard deviation of 41.6 years.